Homework Help: Force of parallel plate capacitor

1. Mar 9, 2016

rgold

1. The problem statement, all variables and given/known data
A parallel-plate capacitor has a plate area of .3m^2 and a plate separation of .1mm. If the charge on ech plate has a magnitude of 5*10^-6C then the force exerted by one plate on the other has what magnitude

2. Relevant equations
Q=E Ԑ A
F=qE

3. The attempt at a solution
so to solve for E i used the first equation: 5*10^-6C=E*(10^-11)(.2) and got E=1.67*10^6 V/m
First i want to make sure that i am moving in the right direction and also do i use 5*10^-6C as my q in the second equation?

2. Mar 9, 2016

rgold

U= (Q^2)/(2c) = (Q^2)/(2ԐA) d and F=(dU)/(dd) = (-Q^2)/ (2ԐA)?
so F= (5*10^-6C)^2/(2 * 10^-11 *.3)?

3. Mar 9, 2016

Staff: Mentor

Your approximation for the value of $\epsilon_o$ is very crude. A good value would be $\epsilon_o = 8.854 \times 10^{-12}~ F/m$, but even $9 \times 10^{-12}~F/m$ would be better.

The appropriate approach to this problem is to find the electric field due to a single charged plate and then determine the force that it exerts on the charge of the other plate. Note that between the plates of a parallel plate capacitor the net field is due to the contribution of the fields from both plates; each plate contributes half the total field.
What does the "(.2)" value represent?

4. Mar 9, 2016

rgold

that was a typo i meant to write .3 m^2. ok i used the 8.85*10^-12 and got 4.71N is this more accurate

5. Mar 9, 2016

Staff: Mentor

That would be a good answer

6. Mar 9, 2016

rgold

thank you very much!